2SO₂+O₂⇒2SO₃ ΔH=-197 kJ
<h3>Further explanation</h3>
Based on the principle of Hess's Law, the change in enthalpy of a reaction will be the same even though it is through several stages or ways
Reaction
2SO₂+O₂⇒2SO₃
Given :
1. S(s)+O₂(g)→SO₂(g) ΔH = -297 kJ
Reverse
SO₂(g) ⇒S(s)+O₂(g ΔH = +297 kJ
(sign change to +) x 2
2SO₂(g) ⇒2S(s)+2O₂(g ΔH = +594 kJ
2.2S(s)+3O2(g)→2SO3(g) ΔH=-791kJ
Add both reactions and remove/subtract the same compound for different sides
1. 2SO₂(g) ⇒2S(s)+2O₂(g) ΔH = +594 kJ
2.2S(s)+3O₂(g)→2SO₃(g) ΔH=-791kJ
--------------------------------------------------------+
2SO₂+O₂⇒2SO₃ ΔH=-197 kJ
